Is 1 = 0.9999......

[dethman]

can't believe so many people voted no.

 

[element]

Originally posted by: TuxDave

Originally posted by: element®

Originally posted by: TuxDave

Originally posted by: Skyclad1uhm1
Btw, for x < 1, what is the largest x possible?

None exists? (according to a philosophy class). They also proposed a question, imagine two line, one rotating and the other still. At one point, the two lines will be parallel. At some time in the future, they will be crossed, where's the first point of contact? Does it exist?

And so... I have no idea why I just mentioned that...

This probably deserves its own thread as it is going off on a tangent...ahhha no pun intended but a punny made nonetheless!

But first of all define a line. Do you mean an inifinitely long ray? This does not physically exist of course. If you mean a finite line of finite length and width then the point at which they touch depends on the instrument you use to determine that point. It depends on our resolution of accuracy in determining when they touch. This is ever changing as technology improves and resolutions increase to see as smaller levels. (Atomic, subatomic...etc...)

I meant an infinitely long ray... it's more of a thought experiment but it cannot be realized in this physical world.

If you mean an infinitely long ray, as we all agree it does not exist, nor ever will exist, seeing as you can never stop "drawing" it, then the whole thought experiment is moot, because two lines that don't exist will never cross nor ever be parallel, so there is nothing to talk about. I wonder how your philosophy professor would take that hehe.

 

[RossGr]

Originally posted by: thraashman

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.

Ok, for this problem to work, 0.999999..... assumes and infinite number of 9's. The problem occurs here
9.9999... - 0.9999... , both assume an infinite number of 9's, the problem is that you cannot subtract and infinite from and infinite.

The reason a problem like this works in Calculus, is that in Calculus, you work with limits. And limits mean a number that is all but achievable. in calculus, 0.99999999...... is the limit as n approaches infinity of 9/(10)^n, 1 is the answer, because 1 is what that number can get infinitely close to, but never quite achieve, therefore in calculus, you assume equivalence because there is no distinguishable difference in value. The best example of this is when someone said 1 - 0.999999999......, and that with an infinite number of 9's, the 1 that should exist after all the 0's, never really exists, and the answer is 0.000000000.... there is a difference, but you can never prove the difference because there are and infinite number of 0's.

So, in other words, 1 != 0.999999.... , but the difference is indistinguishable and for all logical purposes, equivalence is assumed.

Unfortunately you are incorrect .999.... is not a limit, a limit implies that you are approachiing a point. This IS the point.
The algebraic manipulations you have used to "disporve" this fact do not constitute a proof.
This is a proof Please point out the errors.

 

[heliomphalodon]

Originally posted by: thraashman

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.

Ok, for this problem to work, 0.999999..... assumes an infinite number of 9's. The problem occurs here
9.9999... - 0.9999... , both assume an infinite number of 9's, the problem is that you cannot subtract an infinite from an infinite.

The reason a problem like this works in Calculus, is that in Calculus, you work with limits. And limits mean a number that is all but achievable. In calculus, 0.99999999...... is the limit as n approaches infinity of 9/(10)^n, 1 is the answer, because 1 is what that number can get infinitely close to, but never quite achieve, therefore in calculus, you assume equivalence because there is no distinguishable difference in value. The best example of this is when someone said 1 - 0.999999999......, and that with an infinite number of 9's, the 1 that should exist after all the 0's, never really exists, and the answer is 0.000000000.... there is a difference, but you can never prove the difference because there are an infinite number of 0's.

So, in other words, 1 != 0.999999.... , but the difference is indistinguishable and for all logical purposes, equivalence is assumed.

With all due respect, you're completely at sea with this "explanation." I hardly know where to begin. I guess I'll start by noting that thraashman's proof is completely legitimate. There are no "infinites" being subtracted, any more than there are in an operation like 2.000000... - 1.000000...

You're right that limits are involved, but you're confused about what the limit means in this context. The limit appears in the definition of the meaning (value) of the expression "0.999999..." -- the expression is meaningless (valueless) in the absence of the definition. Therein seems to lie the problem that a lot of people have with this question -- they apparently don't understand that without a definition of the meaning of the expression, there is no way to assign any value to it.

If we consider the infinite sequence {0.9, 0.99, 0.999, 0.9999, ... } then it is clear that none of the terms of the sequence is equal to 1. But this is not relevant to the question, because 0.999999... is not a term of the sequence. Instead, the value of the expression "0.999999..." is defined to be the limit of the sequence. This limit is 1, and therefore 0.999999... = 1. While it is true that the successive terms of the sequence do "approach" 1, the expression "0.999999..." doesn't "approach" anything -- it has a single, well-defined value: 1.

Nothing is "assumed" at all. There is no "indistinguishable" difference. To claim otherwise is to fall for Zeno's Paradox.

I'll give one last example. Consider the infinite sum: 1/2 + 1/4 + 1/8 + 1/16 +...
Can you see that this sum is equal to 1 ? If not, it is pointless for you to continue this discussion. If, on the other hand, you do see that the infinite sum is equal to 1, then consider a binary number written as "0.111111..." -- the value of this expression is 1, for exactly the same reason that the infinite sum is equal to 1.

 

[GtPrOjEcTX]

Originally posted by: Skyclad1uhm1

Originally posted by: GtPrOjEcTX
n = .9999....

This means that 10 x n = 9.9999....

Now subtract 10n - n = 9.9999... - .9999...
So 9n = 9.0
n = 1

You've not read a single post in the thread, have you?

just the first few

 

[Haircut]

heliomphalodon, nice post.
Unfortunately there have been many others in a similar vein and the naysayers still don't seem to want to accept the fact.

 

[MadRat]

Its nice we have so many explanations of the .999...=1 that sound like a broken record else one might fall for their broken logic.

I agree that an infinite sequence {0.9, 0.99, 0.999, 0.9999, ... } is not being used by a few to represent .999.... accurately, but it is equal. They both have an endless tail of 9's that add up to no definitive meaning. At the infinite position there is still a "9" value. Btw, gentlemem, to claim there is no infinite position is to deny the very existence of infinity. The value of "infinity" is undefined, but that doesn't mean that we cannot position a value at the infinite position. This is like claiming that a value cannot exist at the nth position because we haven't defined a value for n yet!

You've missed the whole point that a "9" placed at the infinite position will never equal a "10", the very value that turns .999... into a "1".

There can also be an infinitely long string of 0's with a single value of "1" at the infinite position.

Mathematicians derive all of their logic from philosophy. Math does not exist separate from the axioms of philosophy, but rather it depends on them. Anything can be shown with an inaccurate proof which is what this proof does:

Originally posted by: GtPrOjEcTX
n = .9999....
This means that 10 x n = 9.9999....
Now subtract 10n - n = 9.9999... - .9999...
So 9n = 9.0
n = 1

The last "9" becomes shifted from its infinite position with the division by 10 (where GtPrOjEcTX says "10 x n = 9.9999...") and now the last "9" sits in the ("infinite position"-1) position. So now the error lies in his step "10n - n = 9.9999... - .9999..." which should have been "10n - n = 9.999...(.9) - .999...9".* His next step, "9n = 9.0" should have read "9n = 9.0...9". The true value of "n" now becomes "1.0...1". Does the definition of "1.0...1" = ".999...9"? No, it does not. Well, not if you care about the accurate display of the "infinite position".

*NOTE: "...n" represents a value at the infinite position of 10^(-1*"inifinity"). In no way does the value of n equal infinity.

 

[josphII]

There can also be an infinitely long string of 0's with a single value of "1" at the infinite position.

NOT true!!

again, there is no such thing as an infinite long string of numbers with another number after that. infinity, by definition, has no end. so how can you have a number at the nth + 1 position if n = infinity? you cant!

The last "9" becomes shifted from its infinite position by the division by 10 and now the last "9" sits in the ("infinite position"-1) position. So now the error lies in his step "9n = 9.0" which should have read "9n = 9.0...9". The true value of "n" now becomes "1.0...1". Does the definition of "1.0...1" = ".999..."? No, it does not. Well, not if you care about the accurate display of the "infinite position".

NOTE: "...n" represents a value at the infinite position of 10^-("inifinity"). In no way does the value of n equal infinity.

although the "proof" that you are referrring too is in fact not a proof at all, your explaination for why it fails is incorrect. firstly you refer to a last position. there is no last position! a last position simply doesnt exist! MadRat, you completely lack an understanding of the concept of infinity. furthermore you are trying to perform basic mathematical functions on infinity, by claiming that there is an 'infinity - 1 position', which there isnt! infinity - 1 == infinity!!

you are using the same busted logic as those that claim the difference between 1 and 0.999... is 0.000...01, which is not true at all. by the very definition of infinty, the "number" 0.000...01 can not and does not exist.

 

[josphII]

Its nice we have so many explanations of the .999...=1 that sound like a broken record else one might fall for their broken logic.

ive only seen one proof of why 0.999...=1, the one posted by myself (from the mathdr site or whatever its called), and RossGr. its not coincidence that nobody can find fault with this proof without saying something that is simply not true, such as referring to 0.999... as a limit and not a number

 

[MadRat]

Originally posted by: josphII
NOT true!!

again, there is no such thing as an infinite long string of numbers with another number after that. infinity, by definition, has no end. so how can you have a number at the nth + 1 position if n = infinity? you cant!

Again, you confuse infinite position with infinity. These are abstracts accepted in philosophy that our local mathematicians cannot accept. Nonetheless they are valid to apply in the realm of mathematics. It is no different than using an imaginary number i to define two equal numbers that when squared derive a negative value. Mathematically it was not possible so in stepped philosophy. Its well understood that the value of i is not defined but that doesn't mean that mathematicians cannot use it in their proofs.

 

[MadRat]

You can check for the validity of an infinite position if you want. Take .000...1 and times it by 10^"infinity". What is the answer?

 

[josphII]

Originally posted by: MadRat
You can check for the validity of an infinite position if you want. Take .000...1 and times it by 10^"infinity". What is the answer?

there is no such number as 0.000...1, so there is no answer

 

[Haircut]

Originally posted by: MadRat
You can check for the validity of an infinite position if you want. Take .000...1 and times it by 10^"infinity". What is the answer?

We cannot have 0.000...1 to have a number at the end of the decimal necessarily means that the numbers in between have a finite cardinality.

If you are so insistant that we can have an infinite position then what number is at the infinite position of 0.121212121...?

 

[josphII]

Originally posted by: MadRat

Originally posted by: josphII
NOT true!!

again, there is no such thing as an infinite long string of numbers with another number after that. infinity, by definition, has no end. so how can you have a number at the nth + 1 position if n = infinity? you cant!

Again, you confuse infinite position with infinity. These are abstracts accepted in philosophy that our local mathematicians cannot accept. Nonetheless they are valid to apply in the realm of mathematics. It is no different than using an imaginary number i to define two equal numbers that when squared derive a negative value. Mathematically it was not possible so in stepped philosophy. Its well understood that the value of i is not defined but that doesn't mean that mathematicians cannot use it in their proofs.

you are so speaking out of your ass on this one. philosophy? philosophy has absolutely no relevance to this discussion. none whatsoever! furthermore your made up "infinite position" is absoultely not like i. i is very real, i equals sqrt(-1). i appears in many solutions and equations throughout engineering and math

 

[heliomphalodon]

Originally posted by: MadRat
Its nice we have so many explanations of the .999...=1 that sound like a broken record else one might fall for their broken logic.

I agree that an infinite sequence {0.9, 0.99, 0.999, 0.9999, ... } is not being used by a few to represent .999.... accurately, but it is equal. They both have an endless tail of 9's that add up to no definitive meaning. At the infinite position there is still a "9" value. Btw, gentlemem, to claim there is no infinite position is to deny the very existence of infinity. The value of "infinity" is undefined, but that doesn't mean that we cannot position a value at the infinite position. This is like claiming that a value cannot exist at the nth position because we haven't defined a value for n yet!

You've missed the whole point that a "9" placed at the infinite position will never equal a "10", the very value that turns .999... into a "1".

There can also be an infinitely long string of 0's with a single value of "1" at the infinite position.

Mathematicians derive all of their logic from philosophy. Math does not exist separate from the axioms of philosophy, but rather it depends on them. Anything can be shown with an inaccurate proof which is what this proof does:

Originally posted by: GtPrOjEcTX
n = .9999....
This means that 10 x n = 9.9999....
Now subtract 10n - n = 9.9999... - .9999...
So 9n = 9.0
n = 1

The last "9" becomes shifted from its infinite position with the division by 10 (where GtPrOjEcTX says "10 x n = 9.9999...") and now the last "9" sits in the ("infinite position"-1) position. So now the error lies in his step "10n - n = 9.9999... - .9999..." which should have been "10n - n = 9.999...(.9) - .999...9".* His next step, "9n = 9.0" should have read "9n = 9.0...9". The true value of "n" now becomes "1.0...1". Does the definition of "1.0...1" = ".999...9"? No, it does not. Well, not if you care about the accurate display of the "infinite position".

*NOTE: "...n" represents a value at the infinite position of 10^(-1*"inifinity"). In no way does the value of n equal infinity.

<sigh> Sadly, sir, the "broken logic" is yours. By your reference to some "infinite position" it is you who deny the existence of infinity.
To speak, as you do, about

the ("infinite position"-1) position

makes it abundantly clear that you utterly fail to grasp the concept of even the smallest infinity.

It's as I said in my first post to this ridiculous thread -- there can be no debate here, only protestations (from the knowledgeable) and expressions of ignorance. Since none is so blind as he who will not see, I am done with protesting. With this last post, I leave the thread to the willfully ignorant, and those who persist in casting pearls before them. May their bliss be undisturbed.

 

[MadRat]

Originally posted by: Haircut

Originally posted by: MadRat
You can check for the validity of an infinite position if you want. Take .000...1 and times it by 10^"infinity". What is the answer?

We cannot have 0.000...1 to have a number at the end of the decimal necessarily means that the numbers in between have a finite cardinality.

If you are so insistant that we can have an infinite position then what number is at the infinite position of 0.121212121...?

The same fraction that formed it of course.

 

[MadRat]

Originally posted by: josphII
you are so speaking out of your ass on this one. philosophy? philosophy has absolutely no relevance to this discussion. none whatsoever! furthermore your made up "infinite position" is absoultely not like i. i is very real, i equals sqrt(-1). i appears in many solutions and equations throughout engineering and math

The personal attack means you've ran out of argument, eh? Fine, I won. Now apologize like a man.

The term "infinite position" was around long before either of use were born. It is equally as valid as the argument as that of i=sqrt(-1).

 

[josphII]

Originally posted by: MadRat

Originally posted by: Haircut

Originally posted by: MadRat
You can check for the validity of an infinite position if you want. Take .000...1 and times it by 10^"infinity". What is the answer?

We cannot have 0.000...1 to have a number at the end of the decimal necessarily means that the numbers in between have a finite cardinality.

If you are so insistant that we can have an infinite position then what number is at the infinite position of 0.121212121...?

The same fraction that formed it of course.

at the "infinite position" (as only you put it) of this decimal is a fraction?! hilarious! not quite as funny as your "well philosophy says..." b.s. but quite funny none the less

 

[Haircut]

Originally posted by: MadRat

Originally posted by: Haircut

Originally posted by: MadRat
You can check for the validity of an infinite position if you want. Take .000...1 and times it by 10^"infinity". What is the answer?

We cannot have 0.000...1 to have a number at the end of the decimal necessarily means that the numbers in between have a finite cardinality.

If you are so insistant that we can have an infinite position then what number is at the infinite position of 0.121212121...?

The same fraction that formed it of course.

What do you mean by 'the same fraction that formed it'?
If we have a number where the digits after the decimal point consist of alternating 1s and 2s then surely the infinite digit must be either a 1 or a 2.
Which is it?

 

[MadRat]

You can define any value smaller or larger that any other position, that is a constant. Just because a value lay at an infinite position does not mean that the rules do not apply. Constants are always applicable, regardless if you talk about values of significance, null, or infinity.

 

[josphII]

Originally posted by: MadRat

Originally posted by: josphII
you are so speaking out of your ass on this one. philosophy? philosophy has absolutely no relevance to this discussion. none whatsoever! furthermore your made up "infinite position" is absoultely not like i. i is very real, i equals sqrt(-1). i appears in many solutions and equations throughout engineering and math

The personal attack means you've ran out of argument, eh? Fine, I won. Now apologize like a man.

The term "infinite position" was around long before either of use were born. It is equally as valid as the argument as that of i=sqrt(-1).

i=sqrt(-1) is not an argument, its fact

[edit] fact isnt the right word, i=sqrt(-1), by definition, not by argument

 

[MadRat]

Originally posted by: Haircut

Originally posted by: MadRat

Originally posted by: Haircut

Originally posted by: MadRat
You can check for the validity of an infinite position if you want. Take .000...1 and times it by 10^"infinity". What is the answer?

We cannot have 0.000...1 to have a number at the end of the decimal necessarily means that the numbers in between have a finite cardinality.

If you are so insistant that we can have an infinite position then what number is at the infinite position of 0.121212121...?

The same fraction that formed it of course.

What do you mean by 'the same fraction that formed it'?
If we have a number where the digits after the decimal point consist of alternating 1s and 2s then surely the infinite digit must be either a 1 or a 2.
Which is it?

You just insist on a value of 1 or 2, but neither is there. The fraction is there at the infinite position.

It is like assuming that a "3" is present in the "infinite position" of the decimal formed by 1/3. No, 1/3 is not .333... but rather it would be .333...(1/3). Three times .333... leaves an insignificant but nonetheless relevant difference from "1". Three times .333...(1/3) makes a perfect "1". The problem with .333...=1 lay in the insistence to use a base of 10 where a base of 3 belongs.

 

[Haircut]

Originally posted by: MadRat

Originally posted by: Haircut

Originally posted by: MadRat

Originally posted by: Haircut

Originally posted by: MadRat
You can check for the validity of an infinite position if you want. Take .000...1 and times it by 10^"infinity". What is the answer?

We cannot have 0.000...1 to have a number at the end of the decimal necessarily means that the numbers in between have a finite cardinality.

If you are so insistant that we can have an infinite position then what number is at the infinite position of 0.121212121...?

The same fraction that formed it of course.

What do you mean by 'the same fraction that formed it'?
If we have a number where the digits after the decimal point consist of alternating 1s and 2s then surely the infinite digit must be either a 1 or a 2.
Which is it?

You just insist on a value of 1 or 2, but neither is there. The fraction is there at the infinite position.

It is like assuming that a "3" is present in the "infinite position" of the decimal formed by 1/3. No, 1/3 is not .333... but rather it would be .333...(1/3). Three times .333... leaves an insignificant but nonetheless relevant difference from "1". Three times .333...(1/3) makes a perfect "1". The problem with .333...=1 lay in the insistence to use a base of 10 where a base of 3 belongs.

I never mentioned anything about a fraction though. If I choose to have a number whereby we start with 0.12 and then when the last digit is a 2 the next one will be a 1, and conversely when the last digit is a 2 then the next one will be 1.

From my definition of the number I have created there cannot ever be anything other than a 1 or a 2 no matter which digit we choose.
I ask you again, what is the infinite digit?

 

[josphII]

Originally posted by: MadRat

Originally posted by: Haircut

Originally posted by: MadRat

Originally posted by: Haircut

Originally posted by: MadRat
You can check for the validity of an infinite position if you want. Take .000...1 and times it by 10^"infinity". What is the answer?

We cannot have 0.000...1 to have a number at the end of the decimal necessarily means that the numbers in between have a finite cardinality.

If you are so insistant that we can have an infinite position then what number is at the infinite position of 0.121212121...?

The same fraction that formed it of course.

What do you mean by 'the same fraction that formed it'?
If we have a number where the digits after the decimal point consist of alternating 1s and 2s then surely the infinite digit must be either a 1 or a 2.
Which is it?

You just insist on a value of 1 or 2, but neither is there. The fraction is there at the infinite position.

It is like assuming that a "3" is present in the "infinite position" of the decimal formed by 1/3. No, 1/3 is not .333... but rather it would be .333...(1/3). Three times .333... leaves an insignificant but nonetheless relevant difference from "1". Three times .333...(1/3) makes a perfect "1". The problem with .333...=1 lay in the insistence to use a base of 10 where a base of 3 belongs.

lol, a new chapter of "MadRat math" is being written with every post

 

[MadRat]

Originally posted by: Haircut
I ask you again, what is the infinite digit?

I could presuppose you mean something else. I'll let you explain what you want for an answer.